Posted by **Sian** on Tuesday, March 26, 2013 at 9:21am.

An elevator with a man inside, of gross mass 500KG, starts moving upwards with a costant acceleration and acquires a velocity os 2m/s after travelling a distance os 3m.Find the pull in the cables during the accelerated motion.

If the elevator, before stopping, moves with a constant deceleration from a constant velocity of 2m/s and comes to rest in 2 second, calculate the force exerted by the man of 75kg mass on the floor on the elevator.

- Collage Physics -
**black_widow**, Tuesday, March 26, 2013 at 9:51am
The pull of cable is 5.24x10^3N

The force exerted by the man =75N

Is it correct?

- Collage Physics -
**bobpursley**, Tuesday, March 26, 2013 at 10:09am
first, find the acceleration.

Vf^2=Vi^2+2ad

4=0+2a*3

a=2/3 m/s^2

then

force= m(g+a)=500(9.8+2/3) =

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